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2014-09-07T07:26:06+05:30

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Probability of A getting a 6 on throwing one die :  1/6 as there are 6 faces and six numbers.
Probability of A getting at least one 6 on throwing 4 dies =
       = 1 - probability of getting no six at all
       = 1 - 5/6 * 5/6 * 5/6 * 5/6 = 0.5177
Probability of A winning  = 0.5177

probability of B getting a 6 on throwing one die : 1/6

Probability of B getting  no six at all : (5/6)^8 = 0.23256
Probability of B getting 1 six only of 8 dice:
= P(first die = 6, remaining all 1 to 5) + P(2nd die=6, remaining all =1 to 5) + .....
= 8 P(first die = 6, remaining all 1 to 5)
= 8 * (1/6) * (5/6)^7 =  0.3721

Total probability of B getting at least two 6: 1 - 0.23256 - 0.3721 = 0.3853
Probability of B winning = 0.3853 < P(A wins) = 0.5177

A has a higher probability to win the game.

0
how r u so sure
Why do u say it is wrong
The question is who has the greater probability to win. Not, what is the chance of getting six
A will have probability
the answer is not so simple as 1/4 and 2/8. please understand. i am sorry for any inconvenience.